# Prove that a subsequence converges uniformly

$$f_n$$ is a sequence of real functions defined on $$[0,1]$$.

For all n, we have

i) $$f_n(0)=f'_n(0)=0$$

ii) $$f''$$ is continuous, and $$|f''(x)|\leq 1$$ for all $$x \in [0,1]$$

Prove that $${f_n}$$ has a subsequece converging uniformly on $$[0,1]$$

My Solution:

I have proved that $$f_n$$ is pointwise bounded.

Thus, for any $$x\in [0,1]$$, we can find a sequence $${x_k}$$ converging to x and a subsequence of $$f_n$$ (denote this subsequence as $$\{g_n\}$$) converging for every point in $$x_k$$. i.e. given any $$x_k$$, $$\{ g_n(x_k) \}$$ converges as $$n\rightarrow \infty$$

I want to prove that $$\{g_n\}$$ converges uniformly on $$[0,1]$$

I am considering the following:

$$|g_n(x)-g_m(x)|\leq|g_n(x)-g_n(x_k)|+|g_n(x_k)-g_m(x_k)|+|g_m(x_k)-g_m(x)|$$

I think the first part and the third part con be controlled below $$\epsilon$$ by continuous and convergence of $$x_k$$ and the second part can be controlled by pointwise convergence. But when writing the proof, I found that the selection of $$m$$ and $$n$$ depends on $$k$$ , and that the selection of $$k$$ depends on $$m$$ and $$n$$.

Did I ignore anything? Or can I just say, "for any $$\epsilon$$ we can find $$x_k$$, $$N_k$$ such that $$m, n> N_k$$ implies that all conditions hold"?

• do you mean $f_n''$ is continuous and obeys the bound? And can you use Arzela-Ascoli? Nov 11, 2019 at 16:24
• Do you know Arzela-Ascoli?
– zhw.
Nov 11, 2019 at 16:26

Use Arzela-Ascoli Theorem. Since $$|f_n''|\leq 1$$ applying Mean-Value theorem we have, $$\frac{f_n'(x)-f_n'(0)}{x-0}=f_n''(c_{n,x})\implies |f_n'(x)|\leq x|f_n''(c_{n,x})|\leq 1,\forall n\in \Bbb N,\forall x\in [0,1].$$ Now, $$\frac{f_n(x)-f_n(0)}{x-0}=f_n'(d_{n,x})\implies |f_n(x)|\leq x|f_n'(d_{n,x})|\leq 1,\forall x\in [0,1],\forall n\in \Bbb N$$$$\implies \{f_n\}\text{ is uniformly bounded}$$ and $$\forall n\in \Bbb N,\forall x,y\in [0,1]$$ we have, $$\frac{f_n(x)-f_n(y)}{x-y}=f_n'(e_{n,x})\implies |f_n(x)-f_n(y)|\leq |x-y||f_n'(e_{n,x})|\leq |x-y|$$$$\implies \{f_n\}\text{ is equicontinuous family.}$$ So $$\{f_n\}$$ is a equicontinuos family of continuous functions which is uniformly bounded. So by Arzela-Ascoli theorem this seqnce has a uniformly convergent subsequence.